Question

Molar conductivity of a weak acid HA at infinite dilution is $345.8{{S}}.{{c}}{{{m}}^2}.{{mo}}{{{l}}^{ - 1}}$. Calculate the molar conductivity of $0.05{{M}}$ HA solution. Given that $\alpha = 5.8 \times {10^{ - 6}}$

Answer 1

Hint:We know that the molar conductivity is the ability to conduct the ions which are produced by one mole of electrolyte in a given solution. The electrolyte gets dissociated completely at infinite dilution. Since the mobility of ions are different, the molar conductivity is also different.

Complete step by step answer:
It is given that the molar conductivity of HA at infinite dilution, ${\Lambda ^\infty }_{{m}} = 345.8{{S}}.{{c}}{{{m}}^2}.{{mo}}{{{l}}^{ - 1}}$
Concentration of HA solution, ${{c = 0}}{{.05M}}$
Degree of dissociation, $\alpha = 5.8 \times {10^{ - 6}}$
Molar conductivity is represented by \[{\Lambda _{{m}}}\]. We know that the molar conductivity has a relation with specific conductivity as given below:
\[{\Lambda _{{m}}} = \dfrac{{{K}}}{{{c}}}\], where ${{K}}$ is the specific conductivity and ${{c}}$ is the concentration.
When we dilute the solution of electrolyte, molar conductivity is also increased. Degree of dissociation is the part of the total electrolyte existing as ionic form. It is calculated by dividing molar conductivity of a weak electrolyte by molar conductivity of an electrolyte at infinite dilution.
i.e. \[\alpha = \dfrac{{{\Lambda _{{m}}}}}{{{\Lambda ^\infty }_{{m}}}}\], where \[{\Lambda _{{m}}}\] is the molar conductivity of HA and \[{\Lambda ^\infty }_{{m}}\] is the molar conductivity if electrolyte at infinite dilution.
Substituting the values, we get
\[5.8 \times {10^{ - 6}} = \dfrac{{{\Lambda _{{m}}}}}{{345.8{{S}}.{{c}}{{{m}}^2}.{{mo}}{{{l}}^{ - 1}}}}\]
On solving, we get
Molar conductivity of HA, \[{\Lambda _{{m}}} = 5.8 \times {10^{ - 6}} \times 345.8{{S}}.{{c}}{{{m}}^2}.{{mo}}{{{l}}^{ - 1}} = 2.005 \times {10^{ - 3}}{{S}}.{{c}}{{{m}}^2}.{{mo}}{{{l}}^{ - 1}}\]
Thus the molar conductivity of $0.05{{M}}$ HA solution, \[{\Lambda _{{m}}} = 2.005 \times {10^{ - 3}}{{S}}.{{c}}{{{m}}^2}.{{mo}}{{{l}}^{ - 1}}\]

Note:
On dilution, the specific conductivity is decreased while the molar conductivity is increased.
Kohlrausch’s law is the basis for finding the relation between degree of dissociation of an electrolyte, molar conductivity of electrolyte at infinite dilution and molar conductivity of weak acid. On dilution, the volume of the solution increases, but the number of ions increases to a small extent. So the amount of ions per unit volume is decreased.